PKU 3076 Sudoku Dancing Links (DLX) 果然很生猛……
Sudoku
| Time Limit: 10000MS | Memory Limit: 65536K | |
| Total Submissions: 638 | Accepted: 304 |
Description
A Sudoku grid is a 16x16 grid of cells grouped in sixteen 4x4 squares, where some cells are filled with letters from A to P (the first 16 capital letters of the English alphabet), as shown in figure 1a. The game is to fill all the empty grid cells with letters from A to P such that each letter from the grid occurs once only in the line, the column, and the 4x4 square it occupies. The initial content of the grid satisfies the constraints mentioned above and guarantees a unique solution.
Write a Sudoku playing program that reads data sets from a text file.
Write a Sudoku playing program that reads data sets from a text file.
Input
Each data set encodes a grid and contains 16 strings on 16 consecutive lines as shown in figure 2. The i-th string stands for the i-th line of the grid, is 16 characters long, and starts from the first position of the line. String characters are from the set {A,B,…,P,-}, where – (minus) designates empty grid cells. The data sets are separated by single empty lines and terminate with an end of file.
Output
The program prints the solution of the input encoded grids in the same format and order as used for input.
Sample Input
--A----C-----O-I
-J--A-B-P-CGF-H-
--D--F-I-E----P-
-G-EL-H----M-J--
----E----C--G---
-I--K-GA-B---E-J
D-GP--J-F----A--
-E---C-B--DP--O-
E--F-M--D--L-K-A
-C--------O-I-L-
H-P-C--F-A--B---
---G-OD---J----H
K---J----H-A-P-L
--B--P--E--K--A-
-H--B--K--FI-C--
--F---C--D--H-N-
Sample Output
FPAHMJECNLBDKOGI
OJMIANBDPKCGFLHE
LNDKGFOIJEAHMBPC
BGCELKHPOFIMAJDN
MFHBELPOACKJGNID
CILNKDGAHBMOPEFJ
DOGPIHJMFNLECAKB
JEKAFCNBGIDPLHOM
EBOFPMIJDGHLNKCA
NCJDHBAEKMOFIGLP
HMPLCGKFIAENBDJO
AKIGNODLBPJCEFMH
KDEMJIFNCHGAOPBL
GLBCDPMHEONKJIAF
PHNOBALKMJFIDCEG
IAFJOECGLDPBHMNK
Source
Southeastern Europe 2006
题意就是数独啦，数独可以转换成n*n*n行4*n*n列的精确覆盖问题。
每一行对应每一格的一种填法
每一列代表一个状态（类似于八皇后里面的行、列、斜）
9行，每行9个数的状态——》81
9列——》81
9块——》81
81个格，每格只能有一个数——》81
所以列数为n*n*4。
对应每一格每一种填法，必定覆盖某4列（行、列、区域、格子）
用DLX过了一堆数独…………
1
#include
<
stdio.h
>
2 #include < string .h >
3
4 const int n = 16 ;
5 const int len = 4 ;
6
7 const int maxrow = n * n * n; // row, col, value
8 const int maxcol = 4 * n * n; // row, col, block, cell
9
10 struct Node {
11 int row,col,num;
12 Node * left, * right, * up, * down;
13 };
14
15 Node head,row[maxrow + 10 ],col[maxcol + 10 ],node[maxrow * maxcol + 2000 ];
16 int nodenum;
17 int count[maxcol + 10 ];
18 int result[maxrow + 10 ];
19 char temp[ 2000 ];
20
21 void linkinit()
22 {
23 memset(count, 0 , sizeof (count));
24 nodenum = 0 ;
25 head.col = head.row = 0 ;
26 head.right = head.left = head.up = head.down =& head;
27 for ( int i = 0 ;i < maxcol;i ++ ) {
28 col[i].left = head.left;
29 head.left =& col[i];
30 col[i].right =& head;
31 col[i].num = 0 ;
32 col[i].col = i;
33 col[i].up = col[i].down =& col[i];
34 col[i].left -> right =& col[i];col[i].right -> left =& col[i];
35 }
36 for ( int i = 0 ;i < maxrow;i ++ ) {
37 row[i].up = head.up;
38 head.up =& row[i];
39 row[i].row = i;
40 row[i].down =& head;
41 row[i].left = row[i].right =& row[i];
42 row[i].up -> down =& row[i];row[i].down -> up =& row[i];
43 }
44 }
45
46 void link( int r, int c, int num, int coln)
47 {
48 int rown = r * n * n + c * n + num;
49 count[coln] ++ ;
50 node[nodenum].row = rown;node[nodenum].col = coln;
51 node[nodenum].up =& col[coln];
52 node[nodenum].down = col[coln].down;
53 col[coln].down -> up =& node[nodenum];
54 col[coln].down =& node[nodenum];
55 node[nodenum].left =& row[rown];
56 node[nodenum].right = row[rown].right;
57 row[rown].right -> left =& node[nodenum];
58 row[rown].right =& node[nodenum];
59 nodenum ++ ;
60 }
61
62 void removerowhead()
63 {
64 for ( int i = 0 ;i < maxrow;i ++ ) {
65 row[i].left -> right = row[i].right;
66 row[i].right -> left = row[i].left;
67 }
68 }
69
70 void remove( int coln)
71 {
72 col[coln].left -> right = col[coln].right;
73 col[coln].right -> left = col[coln].left;
74 for (Node * curr = col[coln].down;curr !=& col[coln];curr = curr -> down)
75 {
76 for (Node * tmp = curr -> right;tmp != curr;tmp = tmp -> right)
77 {
78 count[tmp -> col] -- ;
79 tmp -> up -> down = tmp -> down;
80 tmp -> down -> up = tmp -> up;
81 }
82 curr -> left -> right = curr -> right;
83 curr -> right -> left = curr -> left;
84 }
85 }
86
87 void resume( int coln)
88 {
89 for (Node * curr = col[coln].down;curr !=& col[coln];curr = curr -> down)
90 {
91 curr -> right -> left = curr;
92 curr -> left -> right = curr;
93 for (Node * tmp = curr -> left;tmp != curr;tmp = tmp -> left)
94 {
95 tmp -> up -> down = tmp;
96 tmp -> down -> up = tmp;
97 count[tmp -> col] ++ ;
98 }
99 }
100 col[coln].left -> right =& col[coln];
101 col[coln].right -> left =& col[coln];
102 }
103
104 bool solve( int k)
105 {
106 if (head.right ==& head) return true ;
107 int min = 2 * n,rec;
108 for (Node * curr = head.right;curr !=& head;curr = curr -> right) {
109 if (min > count[curr -> col]) {
110 min = count[curr -> col];
111 rec = curr -> col;
112 if (min <= 1 ) break ;
113 }
114 }
115 if (min == 0 ) return false ;
116 remove(rec);
117 for (Node * curr = col[rec].down;curr !=& col[rec];curr = curr -> down) {
118 result[k] = curr -> row;
119 curr -> left -> right = curr;
120 for (Node * tmp = curr -> right;tmp != curr;tmp = tmp -> right) {
121 remove(tmp -> col);
122 }
123 curr -> left -> right = curr -> right;
124 if (solve(k + 1 )) return true ;
125 curr -> right -> left = curr;
126 for (Node * tmp = curr -> left;tmp != curr;tmp = tmp -> left)
127 {
128 resume(tmp -> col);
129 }
130 curr -> right -> left = curr -> left;
131 }
132 resume(rec);
133 return false ;
134 }
135
136 void probinit()
137 {
138 int r,c,b;
139 for ( int i = 0 ;i < n * n;i ++ ) {
140 r = i / n;c = i % n;b = (r / len) * len + (c / len);
141 if (temp[i] == ' - ' ) {
142 for ( int j = 0 ;j < n;j ++ ) {
143 link(r,c,j,r * n + j); // row
144 link(r,c,j,n * n + c * n + j); // column
145 link(r,c,j,n * n * 2 + b * n + j); // block
146 link(r,c,j,n * n * 3 + i); // cell
147 }
148 }
149 else {
150 int j = temp[i] - ' A ' ;
151 link(r,c,j,r * n + j); // row
152 link(r,c,j,n * n + c * n + j); // column
153 link(r,c,j,n * n * 2 + b * n + j); // block
154 link(r,c,j,n * n * 3 + i); // cell
155 }
156 }
157 }
158
159 void printres()
160 {
161 int r,c,num;
162 int numb[n * n + 10 ] = { 0 };
163 for ( int i = 0 ;i < n * n;i ++ )
164 {
165 r = result[i] / (n * n);
166 c = (result[i] - r * n * n) / n;
167 num = result[i] % n;
168 numb[r * n + c] = num + 1 ;
169 }
170 for ( int i = 0 ;i < n * n;i ++ ) {
171 printf( " %c " ,numb[i] + ' A ' - 1 );
172 if ((i + 1 ) % n == 0 ) printf( " \n " );
173 }
174 printf( " \n " );
175 }
176
177 int main()
178 {
179 char tmpp[ 100 ];
180 while (scanf( " %s " ,tmpp) != EOF)
181 {
182 memset(temp, 0 , sizeof (temp));
183 strcat(temp,tmpp);
184 for ( int i = 1 ;i < n;i ++ ) {
185 scanf( " %s " ,tmpp);
186 strcat(temp,tmpp);
187 }
188 linkinit();
189 probinit();
190 removerowhead();
191 solve( 0 );
192 printres();
193 }
194 return 0 ;
195 }
196
2 #include < string .h >
3
4 const int n = 16 ;
5 const int len = 4 ;
6
7 const int maxrow = n * n * n; // row, col, value
8 const int maxcol = 4 * n * n; // row, col, block, cell
9
10 struct Node {
11 int row,col,num;
12 Node * left, * right, * up, * down;
13 };
14
15 Node head,row[maxrow + 10 ],col[maxcol + 10 ],node[maxrow * maxcol + 2000 ];
16 int nodenum;
17 int count[maxcol + 10 ];
18 int result[maxrow + 10 ];
19 char temp[ 2000 ];
20
21 void linkinit()
22 {
23 memset(count, 0 , sizeof (count));
24 nodenum = 0 ;
25 head.col = head.row = 0 ;
26 head.right = head.left = head.up = head.down =& head;
27 for ( int i = 0 ;i < maxcol;i ++ ) {
28 col[i].left = head.left;
29 head.left =& col[i];
30 col[i].right =& head;
31 col[i].num = 0 ;
32 col[i].col = i;
33 col[i].up = col[i].down =& col[i];
34 col[i].left -> right =& col[i];col[i].right -> left =& col[i];
35 }
36 for ( int i = 0 ;i < maxrow;i ++ ) {
37 row[i].up = head.up;
38 head.up =& row[i];
39 row[i].row = i;
40 row[i].down =& head;
41 row[i].left = row[i].right =& row[i];
42 row[i].up -> down =& row[i];row[i].down -> up =& row[i];
43 }
44 }
45
46 void link( int r, int c, int num, int coln)
47 {
48 int rown = r * n * n + c * n + num;
49 count[coln] ++ ;
50 node[nodenum].row = rown;node[nodenum].col = coln;
51 node[nodenum].up =& col[coln];
52 node[nodenum].down = col[coln].down;
53 col[coln].down -> up =& node[nodenum];
54 col[coln].down =& node[nodenum];
55 node[nodenum].left =& row[rown];
56 node[nodenum].right = row[rown].right;
57 row[rown].right -> left =& node[nodenum];
58 row[rown].right =& node[nodenum];
59 nodenum ++ ;
60 }
61
62 void removerowhead()
63 {
64 for ( int i = 0 ;i < maxrow;i ++ ) {
65 row[i].left -> right = row[i].right;
66 row[i].right -> left = row[i].left;
67 }
68 }
69
70 void remove( int coln)
71 {
72 col[coln].left -> right = col[coln].right;
73 col[coln].right -> left = col[coln].left;
74 for (Node * curr = col[coln].down;curr !=& col[coln];curr = curr -> down)
75 {
76 for (Node * tmp = curr -> right;tmp != curr;tmp = tmp -> right)
77 {
78 count[tmp -> col] -- ;
79 tmp -> up -> down = tmp -> down;
80 tmp -> down -> up = tmp -> up;
81 }
82 curr -> left -> right = curr -> right;
83 curr -> right -> left = curr -> left;
84 }
85 }
86
87 void resume( int coln)
88 {
89 for (Node * curr = col[coln].down;curr !=& col[coln];curr = curr -> down)
90 {
91 curr -> right -> left = curr;
92 curr -> left -> right = curr;
93 for (Node * tmp = curr -> left;tmp != curr;tmp = tmp -> left)
94 {
95 tmp -> up -> down = tmp;
96 tmp -> down -> up = tmp;
97 count[tmp -> col] ++ ;
98 }
99 }
100 col[coln].left -> right =& col[coln];
101 col[coln].right -> left =& col[coln];
102 }
103
104 bool solve( int k)
105 {
106 if (head.right ==& head) return true ;
107 int min = 2 * n,rec;
108 for (Node * curr = head.right;curr !=& head;curr = curr -> right) {
109 if (min > count[curr -> col]) {
110 min = count[curr -> col];
111 rec = curr -> col;
112 if (min <= 1 ) break ;
113 }
114 }
115 if (min == 0 ) return false ;
116 remove(rec);
117 for (Node * curr = col[rec].down;curr !=& col[rec];curr = curr -> down) {
118 result[k] = curr -> row;
119 curr -> left -> right = curr;
120 for (Node * tmp = curr -> right;tmp != curr;tmp = tmp -> right) {
121 remove(tmp -> col);
122 }
123 curr -> left -> right = curr -> right;
124 if (solve(k + 1 )) return true ;
125 curr -> right -> left = curr;
126 for (Node * tmp = curr -> left;tmp != curr;tmp = tmp -> left)
127 {
128 resume(tmp -> col);
129 }
130 curr -> right -> left = curr -> left;
131 }
132 resume(rec);
133 return false ;
134 }
135
136 void probinit()
137 {
138 int r,c,b;
139 for ( int i = 0 ;i < n * n;i ++ ) {
140 r = i / n;c = i % n;b = (r / len) * len + (c / len);
141 if (temp[i] == ' - ' ) {
142 for ( int j = 0 ;j < n;j ++ ) {
143 link(r,c,j,r * n + j); // row
144 link(r,c,j,n * n + c * n + j); // column
145 link(r,c,j,n * n * 2 + b * n + j); // block
146 link(r,c,j,n * n * 3 + i); // cell
147 }
148 }
149 else {
150 int j = temp[i] - ' A ' ;
151 link(r,c,j,r * n + j); // row
152 link(r,c,j,n * n + c * n + j); // column
153 link(r,c,j,n * n * 2 + b * n + j); // block
154 link(r,c,j,n * n * 3 + i); // cell
155 }
156 }
157 }
158
159 void printres()
160 {
161 int r,c,num;
162 int numb[n * n + 10 ] = { 0 };
163 for ( int i = 0 ;i < n * n;i ++ )
164 {
165 r = result[i] / (n * n);
166 c = (result[i] - r * n * n) / n;
167 num = result[i] % n;
168 numb[r * n + c] = num + 1 ;
169 }
170 for ( int i = 0 ;i < n * n;i ++ ) {
171 printf( " %c " ,numb[i] + ' A ' - 1 );
172 if ((i + 1 ) % n == 0 ) printf( " \n " );
173 }
174 printf( " \n " );
175 }
176
177 int main()
178 {
179 char tmpp[ 100 ];
180 while (scanf( " %s " ,tmpp) != EOF)
181 {
182 memset(temp, 0 , sizeof (temp));
183 strcat(temp,tmpp);
184 for ( int i = 1 ;i < n;i ++ ) {
185 scanf( " %s " ,tmpp);
186 strcat(temp,tmpp);
187 }
188 linkinit();
189 probinit();
190 removerowhead();
191 solve( 0 );
192 printres();
193 }
194 return 0 ;
195 }
196